Comment by ExUtumno
10 years ago
Thanks!
I'm not sure, but I think that Penrose tilesets are what I call "easy": you can't run into a situation where you can't place a new tile. It would be great if someone here could confirm or deny this.
So if this is the case, then Penrose tilesets are not interesting to WFC, because you can produce arbitrary tilings with much simpler algorithms.
Right now though WFC is only working with square tiles, but it's not hard to generalize it to arbitrary shapes. Paul F. Harrison made a tiling program that supports hex tiles: http://logarithmic.net/pfh/ghost-diagrams See also the relevant paragraph in the readme (just search the word "easy").
Agreed. The best way to find 'complexity' and perhaps aperiodicity is by using contradictory rules.
source: my own quasicrystal simulations (http://www.nature.com/nmat/journal/v14/n1/extref/nmat4152-s2...)
Penrose tiles are "easy" on an unbounded canvas, but I'm pretty sure they're a 100%-probable "contradiction" (because they're aperiodic) on a bounded toroidal canvas.
I think you're right about Penrose tiles being easy. One could, however, potentially use this with polygonal tiles to find their Heesch numbers and maybe even make some advancements in solving some of the many unsolved problems associated with Heesch tiling.
Maybe it could be interesting to place Penrose tiles with the simple algorithm, but color them with your algorithm just like you're currently coloring squares.
So basically make a not-easy tileset with the shapes of Penrose tiles. Yes, this could be interesting.
You need a backtracking algorithm to do Penrose tiles, so yeah, you can get stuck.